Thinking about the problem of the existence of odd perfect numbers, I elaborated the following ideas. I post them here for you to check if they are correct, or point out the errors.
We define the sum of divisors function $\sigma\left(n\right)$ as the sum of all the proper divisors of some positive integer $n$. It is not difficult to show that $\sigma\left(n\right)$ has the following properties:
• For some prime number $p$, noting that the sum of divisors function $\sigma\left(p^{\alpha}\right)$ is a geometric progression; we have that $$\sigma\left(p^{\alpha}\right)=1+p+p^{2}+...+p^{\alpha}=\frac{p^{\alpha+1}-1}{p-1}$$
• The sum of divisors function is multiplicative: if two positive integers $m$ and $n$ are relatively prime, then $$\sigma\left(mn\right)=\sigma\left(m\right)\sigma\left(n\right)$$
We define a perfect number $P$ as some positive integer such that $\sigma\left(P\right)=2P$.
Euler's Odd perfect number theorem states that any odd perfect number $P$ must be of the form $P=m^{2}q^{\alpha}$, with $q$ prime, $\alpha\geq1$ and $\gcd\left(m^{2},q\right)=1$.
If $P=m^{2}q^{\alpha}$ and $P$ is some perfect number, then
$$\sigma\left(m^{2}q^{\alpha}\right)=2m^{2}q^{\alpha}$$
As by Euler's odd perfect number theorem we have that $\gcd\left(m^{2},q\right)=1$, then
$$\sigma\left(m^{2}q^{\alpha}\right)=\sigma\left(m^{2}\right)\sigma\left(q^{\alpha}\right)$$
And substituting,
$$2m^{2}q^{\alpha}=\sigma\left(m^{2}\right)\sigma\left(q^{\alpha}\right)$$
As $\sigma\left(q^{\alpha}\right)=\frac{q^{\alpha+1}-1}{q-1}$, substituting we have that
$$2m^{2}q^{\alpha}=\sigma\left(m^{2}\right)\left(\frac{q^{\alpha+1}-1}{q-1}\right) (1)$$
Noting that $\gcd\left(q^{\alpha},\left(\frac{q^{\alpha+1}-1}{q-1}\right)\right)=1$, it follows that $q^{\alpha}$ must divide $\sigma\left(m^{2}\right)$.
Other hand, it is easy to prove that for any integer $n$, $n^k<\frac{n^{k+1}-1}{n-1}<2n^k$. Therefore, $q^{\alpha}<\left(\frac{q^{\alpha+1}-1}{q-1}\right)<2q^{\alpha}$, and for $(1)$ to hold, necessarily $m^2<\sigma\left(m^{2}\right)<2m^2$.
Now, we have two possible cases:
Case (a): $q^{\alpha}>m^2$
In this case, we have that $2q^{\alpha}>2m^2>\sigma\left(m^{2}\right)$; therefore, as we have that $q^{\alpha}$ must divide $\sigma\left(m^{2}\right)$, we get that necessarily $q^{\alpha}=\sigma\left(m^{2}\right)$.
Substituting, we have that $$q^{\alpha}\left(\frac{q^{\alpha+1}-1}{q-1}\right)=2m^{2}q^{\alpha}$$ $$\frac{q^{\alpha+1}-1}{q-1}=2m^{2}$$
It can be proved (see, for instance, this MO post) that the equation $\frac{q^{\alpha+1}-1}{q-1}=2m^{2}$ has solution only for $\alpha+1 \leq 2$. Thus, in this case we get that necessarily $\alpha = 1$, and some odd perfect number $P$ such that $q^{\alpha}>m^2$ can exist only if $P=qm^2$.
However, $\sigma\left(q\right)\sigma\left(m^{2}\right)=(1+q)\sigma\left(m^{2}\right)$. As we have that $m^2<\sigma\left(m^{2}\right)<2m^2$, then $P=qm^2$ is an abundant number, and can not be perfect.
Case (b): $q^{\alpha}<m^2$
In this case, $\sigma\left(m^{2}\right)=q^{\alpha}*s$, where $s$ is some odd integer greater than $1$. Substituting, we have that $$q^{\alpha}s\left(\frac{q^{\alpha+1}-1}{q-1}\right)=2m^{2}q^{\alpha}$$ $$s\left(\frac{q^{\alpha+1}-1}{q-1}\right)=2m^{2}$$ $$\frac{q^{\alpha+1}-1}{q-1}=\frac{2m^{2}}{s}$$
As $\frac{q^{\alpha+1}-1}{q-1}$ is some integer, and $s$ is odd, it follows that $s$ divides $m^2$, and subsequently $s$ is some divisor of $P$. As $\gcd(q^{\alpha},m^2)=1$, it follows that $\gcd(q,s)=1$.
Before dealing further with case (b), I would be grateful if someone could confirm that I can discard case (a) based on the proof exposed.
Thanks in advance!
Comment converted into an answer, so that this question does not remain in the unanswered queue
You have basically rediscovered my results in The Abundancy Index of Divisors of Odd Perfect Numbers (Theorem 5, page 3), so this is not new.